Let $R$ be a ring(not necessary have "1") and let $I,J$ be ideals of $R$ such that $I+J=R$. I want to prove that there is a $x\in R$ such that
$$x\equiv r ({\rm mod} I) \quad x \equiv s ({\rm mod} J) \quad \mbox{for any} ~~~r,s\in R$$

You're not in vector space.
The hypothesis that $I$ and $J$ are stranger ideals i.e. they are such that $I+J=R$, does not imply that $\forall x \in R, \exists x_i \in I, x_j \in J$ such that $x=x_i+x_j$. What it [only] implies is $\exists a_i \in I, b_j \in J$ such that $a_i+b_j=1$.
If you multiply both sides by $x$ you end up in $I$ or $J$ because of the very definition of what an ideal is and because of the property $I+J=R$.

Let $a_i + b_j = 1$. Multiply both sides by $x$, you get $xa_i+xb_j=x$. Now $xa_i$ is in $I$ and $xb_j$ is in $J$, because $I$ and $J$ are ideals. That is, for every $x\in R$ you get $x_i\in I$ and $x_j\in J$ such that $x_i+x_j=x$.
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Gregor BrunsSep 25 '12 at 10:38